Termination Proof using AProVE

Term Rewriting System R:
[x]
active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

ACTIVE(f(x)) -> F(f(x))
CHK(no(f(x))) -> F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
CHK(no(f(x))) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(X)))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(f(X))))))))
CHK(no(f(x))) -> F(f(f(f(f(f(f(X)))))))
CHK(no(f(x))) -> F(f(f(f(f(f(X))))))
CHK(no(f(x))) -> F(f(f(f(f(X)))))
CHK(no(f(x))) -> F(f(f(f(X))))
CHK(no(f(x))) -> F(f(f(X)))
CHK(no(f(x))) -> F(f(X))
CHK(no(f(x))) -> F(X)
CHK(no(c)) -> ACTIVE(c)
MAT(f(x), f(y)) -> F(mat(x, y))
MAT(f(x), f(y)) -> MAT(x, y)
F(active(x)) -> ACTIVE(f(x))
F(active(x)) -> F(x)
F(no(x)) -> F(x)
F(mark(x)) -> F(x)
TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
TP(mark(x)) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
TP(mark(x)) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(f(X)))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(f(X))))))))
TP(mark(x)) -> F(f(f(f(f(f(f(X)))))))
TP(mark(x)) -> F(f(f(f(f(f(X))))))
TP(mark(x)) -> F(f(f(f(f(X)))))
TP(mark(x)) -> F(f(f(f(X))))
TP(mark(x)) -> F(f(f(X)))
TP(mark(x)) -> F(f(X))
TP(mark(x)) -> F(X)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))
ACTIVE(f(x)) -> F(f(x))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The following dependency pairs can be strictly oriented:

F(active(x)) -> F(x)
F(active(x)) -> ACTIVE(f(x))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(f(x)) -> mark(f(f(x)))
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(active(x₁)) = 1 + x₁

POL(ACTIVE(x₁)) = x₁

POL(no(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(f(x₁)) = x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pairs:

F(mark(x)) -> F(x)
F(no(x)) -> F(x)
ACTIVE(f(x)) -> F(f(x))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳DGraph

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pairs:

F(no(x)) -> F(x)
F(mark(x)) -> F(x)

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The following dependency pair can be strictly oriented:

F(no(x)) -> F(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(no(x₁)) = 1 + x₁

POL(mark(x₁)) = x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳DGraph

...

               →DP Problem 6

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

F(mark(x)) -> F(x)

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The following dependency pair can be strictly oriented:

F(mark(x)) -> F(x)

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(mark(x₁)) = 1 + x₁

POL(F(x₁)) = x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 4

             ↳DGraph

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The following dependency pair can be strictly oriented:

CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

active(f(x)) -> mark(f(f(x)))
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(active(x₁)) = 0

POL(c) = 0

POL(X) = 0

POL(no(x₁)) = x₁

POL(mark(x₁)) = 0

POL(y) = 1

POL(f(x₁)) = 1 + x₁

POL(CHK(x₁)) = 1 + x₁

POL(mat(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pair:

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

two new Dependency Pairs are created:

TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
TP(mark(c)) -> TP(chk(no(c)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 9

             ↳Narrowing Transformation

Dependency Pairs:

TP(mark(c)) -> TP(chk(no(c)))
TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 9

             ↳Nar

...

               →DP Problem 10

                 ↳Polynomial Ordering

Dependency Pair:

TP(mark(c)) -> TP(chk(no(c)))

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The following dependency pair can be strictly oriented:

TP(mark(c)) -> TP(chk(no(c)))

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
active(f(x)) -> mark(f(f(x)))
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(chk(x₁)) = 0

POL(active(x₁)) = 0

POL(c) = 1

POL(X) = 0

POL(no(x₁)) = 0

POL(TP(x₁)) = x₁

POL(mark(x₁)) = x₁

POL(y) = 0

POL(f(x₁)) = 0

POL(mat(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Nar

           →DP Problem 9

             ↳Nar

...

               →DP Problem 11

                 ↳Dependency Graph

Dependency Pair:

Rules:

active(f(x)) -> mark(f(f(x)))
chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
chk(no(c)) -> active(c)
mat(f(x), f(y)) -> f(mat(x, y))
mat(f(x), c) -> no(c)
f(active(x)) -> active(f(x))
f(no(x)) -> no(f(x))
f(mark(x)) -> mark(f(x))
tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes

POL(active(x₁))	= 1 + x₁
POL(ACTIVE(x₁))	= x₁
POL(no(x₁))	= x₁
POL(mark(x₁))	= x₁
POL(f(x₁))	= x₁
POL(F(x₁))	= x₁

POL(active(x₁))	= 0
POL(c)	= 0
POL(X)	= 0
POL(no(x₁))	= x₁
POL(mark(x₁))	= 0
POL(y)	= 1
POL(f(x₁))	= 1 + x₁
POL(CHK(x₁))	= 1 + x₁
POL(mat(x₁, x₂))	= x₂